Question

A body falling from rest describes distances s1, s2, and s3 in the first second and third seconds of its motion. Find ratio of s1 : s2 : s3

Answer 1

Explanation:

1:2:3

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Answer 2

Answer:

S=ut+12gt2S=ut+12gt2, u=0u=0

S=12gt2S=12gt2

Here s is displacement.

Distance dropped in 1st1st second, S1=12g×(1)2=12g×1S1=12g×(1)2=12g×1

Distance dropped in 2nd2nd second, S2=12g×(2)2−12g×(1)2=12g×3S2=12g×(2)2−12g×(1)2=12g×3

Distance dropped in 3rd3rd second, S3=12g×(3)2−12g×(2)2=12g×5S3=12g×(3)2−12g×(2)2=12g×5

S1:S2:S3=1:3:5